Cornelsen Fokus Physik Lösungen

In order to safely conduct any experiment in the laboratory, it is crucial that you

vekshin1

Lab Safety Rules:

Report all accidents, injuries, and breakage of glass or equipment to instructor immediately. Keep pathways clear by placing extra items (books, bags, etc.) on the shelves or under the work tables. If under the tables, make sure that these items can not be stepped on. Long hair (chin-length or longer) must be tied back to avoid catching fire. Wear sensible clothing including footwear. Loose clothing should be secured so they do not get caught in a flame or chemicals.Work quietly — know what you are doing by reading the assigned experiment before you start to work. Pay close attention to any cautions described in the laboratory exercises Do not taste or smell chemicals. Wear safety goggles to protect your eyes when heating substances, dissecting, etc. Do not attempt to change the position of glass tubing in a stopper. Never point a test tube being heated at another student or yourself. Never look into a test tube while you are heating it.Unauthorized experiments or procedures must not be attempted.Keep solids out of the sink. Leave your work station clean and in good order before leaving the laboratory. Do not lean, hang over or sit on the laboratory tables. Do not leave your assigned laboratory station without permission of the teacher. Learn the location of the fire extinguisher, eye wash station, first aid kit and safety shower. Fooling around or "horse play" in the laboratory is absolutely forbidden. Students found in violation of this safety rule will be barred from participating in future labs and could result in suspension. Anyone wearing acrylic nails will not be allowed to work with matches, lighted splints, Bunsen burners, etc. Do not lift any solutions, glassware or other types of apparatus above eye level. Follow all instructions given by your teacher.Learn how to transport all materials and equipment safely. No eating or drinking in the lab at any time!

3 0

3 years ago

5. Graph A below plots a race car's speed for 5 seconds. The car's rate of acceleration is 6 m/s^2

Georgia [21]

Answer:

The answer is below

Explanation:

We are to check if the statement is true of false. If it is false, we correct the statement.

Solution:

Acceleration is the time rate of change of velocity. It is the ratio of the change in velocity to the change in time. The acceleration can be gotten from a velocity time graph by finding the slope of the graph.

The x coordinate represent the time and the y coordinate velocity.

5) Graph A passes through the point (0, 0) and (4, 24). Therefore the acceleration (slope) is:

Acceleration = $\frac{24-0}{4-0}=6\ m/s^2$

This is correct.

6) Graph B is a straight line of 12 m/s. It passes through (0, 12) and (4, 12). Hence:

Acceleration = $\frac{12-12}{4-0}=0\ m/s^2$

This is false.

Therefore the acceleration of graph B is 0 m/s².

8 0

3 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

Series circuits are characterized by the fact that there is a single pathway by which charge can travel. True or false?

Ulleksa [173]

I Think Its True My Dude Or Dudette
.
Hope this helps
.
Zane

6 0

3 years ago

Read 2 more answers

Monochromatic light falling on two very narrow slits 0.048mm apart. Successive fringes on a screen 5.00m away are 6.5cm apart ne

tino4ka555 [31]

Answer:

λ = 5.85 x 10⁻⁷ m = 585 nm

f = 5.13 x 10¹⁴ Hz

Explanation:

We will use Young's Double Slit Experiment's Formula here:

$Y = \frac{\lambda L}{d}\\\\\lambda = \frac{Yd}{L}$

where,

λ = wavelength = ?

Y = Fringe Spacing = 6.5 cm = 0.065 m

d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m

L = screen distance = 5 m

Therefore,

$\lambda = \frac{(0.065\ m)(4.8\ x\ 10^{-5}\ m)}{5\ m}$

λ = 5.85 x 10⁻⁷ m = 585 nm

Now, the frequency can be given as:

$f = \frac{c}{\lambda}$

where,

f = frequency = ?

c = speed of light = 3 x 10⁸ m/s

Therefore,

$f = \frac{3\ x\ 10^8\ m/s}{5.85\ x\ 10^{-7}\ m}\\\\$

f = 5.13 x 10¹⁴ Hz

5 0

3 years ago